21牛客9C - Cells(LGV引理)

题目

C-Cells_2021牛客暑期多校训练营9

题解

先放一个LGV引理的链接在这里。

主要讲讲题解中这里的推导

每次从后往前，后一列减去\(t\)倍前一列可得

\[(a_i+1)\prod_{k=2}^{j+1}{(a_i+k)}-t(a_i+1)\prod_{k=2}^{j}{(a_i+k)}=(a_i+1)(a_i+j+1-t)\prod_{k=2}^j{(a_i+k)} \]

令\(t=j\)，可得

\[(a_i+1)^2\prod_{k=2}^j{(a_i+k)} \]

重复这个过程，直到连乘的项消去，剩余第\(j\)列就是\((a_i+1)^j\)。

这个是范德蒙矩阵的形式，具体化简过程百度或自己推。

最后要求\(\prod_{1\le i < j \le n}{(a_j-a_i)}\)​​，直接卷积求出所有差值的个数然后快速幂即可。由于相同差值至多出现1e5次，可以直接ntt。注意多项式乘法最后度数要乘2，这样卷积出来的结果才是正确的。

#include <bits/stdc++.h>

#define endl '\n'
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define mp make_pair
#define seteps(N) fixed << setprecision(N) 
typedef long long ll;

using namespace std;
/*-----------------------------------------------------------------*/

ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
#define INF 0x3f3f3f3f

const int N = 3e6;
const int M = 998244353;
const double eps = 1e-5;

int rev[N];

inline ll qpow(ll a, ll b, ll m) {
    ll res = 1;
    while(b) {
        if(b & 1) res = (res * a) % m;
        a = (a * a) % m;
        b = b >> 1;
    }
    return res;
}

void change(ll y[], int len) {
    for(int i = 0; i < len; ++i) {
        rev[i] = rev[i >> 1] >> 1;
        if(i & 1) {
            rev[i] |= len >> 1;
        }
    }
    for(int i = 0; i < len; ++i) {
        if(i < rev[i]) {
            swap(y[i], y[rev[i]]);
        }
    }
    return;
}

void fft(ll y[], int len, int on) {
    change(y, len);
    for(int h = 2; h <= len; h <<= 1) {
        ll gn = qpow(3, (M - 1) / h, M);
        if(on == -1) gn = qpow(gn, M - 2, M);
        for(int j = 0; j < len; j += h) {
            ll g = 1;
            for(int k = j; k < j + h / 2; k++) {
              ll u = y[k];
              ll t = g * y[k + h / 2] % M;
              y[k] = (u + t) % M;
              y[k + h / 2] = (u - t + M) % M;
              g = g * gn % M;
            }
        }
    }
    if(on == -1) {
        ll inv = qpow(len, M - 2, M);
        for(int i = 0; i < len; i++) {
            y[i] = y[i] * inv % M;
        }
    }
}

int get(int x) {
    int res = 1;
    while(res < x) {
        res <<= 1;
    }
    return res;
}

ll f1[N], f2[N];

int main() {
    IOS;
    int n;
    cin >> n;
    int mx = 1000000;
    int up = 0;
    ll rj = 1, prod = 1, tj = 1;
    for(int i = 1; i <= n; i++) {
        int x;
        cin >> x;
        f1[x] = 1;
        f2[mx - x] = 1;
        prod = prod * (x + 1) % M;
        tj = tj * i % M;
        rj = rj * tj % M;
        up = x;
    }
    rj = qpow(rj, M - 2, M);
    int len = get(2 * mx + 1);
    fft(f1, len, 1);
    fft(f2, len, 1);
    for(int i = 0; i < len; i++) f1[i] = f1[i] * f2[i] % M;
    fft(f1, len, -1);
    ll ans = 1;
    for(int i = 1; i <= mx; i++) {
        ans = ans * qpow(i, f1[mx - i], M) % M;
    }
    ans = ans * rj % M * prod % M;
    cout << ans << endl;
}

标签：f1,21,int,9C,rev,len,牛客,prod,rj
来源： https://www.cnblogs.com/limil/p/15144595.html